Question

Let the length of the focal chord \( PQ \) of the parabola \( y^2 = 12x \) be 15 units. If the distance of \( PQ \) from the origin is \( p \), then \( 10p^2 \) is equal to _____

Answer 1

Correct Answer: 72

Given the parabola: 
\[ y^2 = 12x \] 

The length of a focal chord is given by: 
\[ \text{Length of focal chord} = 4a \csc^2 \theta = 15 \] 
For this parabola, \(4a = 12\), so: \[ 12 \csc^2 \theta = 15 \] 

Solving for \(\csc^2 \theta\): 
\[ \csc^2 \theta = \frac{15}{12} = \frac{5}{4} \]
Thus: \[ \sin^2 \theta = \frac{4}{5} \] 
Using the trigonometric identity \(\tan^2 \theta = \frac{\sin^2 \theta}{1 - \sin^2 \theta}\):
\[ \tan^2 \theta = \frac{\frac{4}{5}}{1 - \frac{4}{5}} = 4 \implies \tan \theta = 2 \] 
The slope of the focal chord PQ is \(\tan \theta = 2\). 

The equation of the chord passing through the focus \((3, 0)\) is given by: \[ y - 0 = 2(x - 3) \] 

Simplifying: \[ y = 2x - 6 \implies 2x - y - 6 = 0 \] 
To find the perpendicular distance of this line from the origin \((0, 0)\), use the formula:
\[ p = \frac{|2 \times 0 - 0 - 6|}{\sqrt{2^2 + (-1)^2}} = \frac{6}{\sqrt{5}} \] 

Calculating \(10p^2\): 
\[ 10p^2 = 10 \left(\frac{6}{\sqrt{5}}\right)^2 = 10 \times \frac{36}{5} = 72 \]